Difference between revisions of "1978 AHSME Problems/Problem 1"
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==Solution 1== | ==Solution 1== | ||
By guessing and checking, 2 works. | By guessing and checking, 2 works. | ||
| − | <math>\frac{2}{x} = \boxed{\textbf{( | + | <math>\frac{2}{x} = \boxed{\textbf{(B) }1}</math> |
| + | ~awin | ||
| + | |||
| + | ==Solution 2== | ||
| + | Multiplying each side by <math>x^2</math>, we get <math>x^2-4x+4 = 0</math>. Factoring, we get <math>(x-2)(x-2) = 0</math>. Therefore, <math>x = 2</math>. | ||
| + | <math>\frac{2}{x} = \boxed{\textbf{(B) }1}</math> | ||
| + | ~awin | ||
| + | |||
| + | ==Solution 3== | ||
| + | Directly factoring, we get <math>(1-\frac{2}{x})^2 = 0</math>. Thus <math>\frac{2}{x}</math> must equal <math>\boxed{\textbf{(B) }1}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1978|num-b=0|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 10:59, 13 February 2021
Problem 1
If 
, then 
equals
Solution 1
By guessing and checking, 2 works.
~awin
Solution 2
Multiplying each side by 
, we get 
. Factoring, we get 
. Therefore, 
.
~awin
Solution 3
Directly factoring, we get 
. Thus must equal 
See Also
| 1978 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 0 |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
